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\centerline{Details of the fermionic force for overlap fermions}
\rightline{Rajamani Narayanan}
\rightline{2/19/99}
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Our starting point is the pseudo-fermion action
$$S_p = \phi^\dagger_+ \bigl[H^2\bigr]^{-1} \phi_+$$
where
$$H^2= (1-\mu^2) \bigl[ {1\over 2} + {1\over 4}(1+\gamma_5)\epsilon(H_w)
+{\mu^2\over 1-\mu^2} \bigr]$$
In the pseudo-fermion action, the pesudo-fermions, $\phi_+$ have 
a positive chirality and should be thought of as having only
two spin components in 4D ( one spin component in 2D).

Let is compute the change in this action with respect to the gauge
field $U$:
$$\eqalign{ {\delta S_p\over \delta U} & = 
-\phi^\dagger_+ \bigl[H^2\bigr]^{-1} {\delta H^2\over \delta U}
\bigl[H^2\bigr]^{-1} \phi_+\cr
& = -\psi^\dagger_+ {\delta H^2\over \delta U} \psi_+ \cr
& = -{1\over 2} (1-\mu^2) \psi^\dagger_+ {\delta \epsilon(H_w)\over \delta U} 
\psi_+ \cr}$$
We have used the notation:
$$\bigl[H^2\bigr]^{-1} \phi_+ = \psi_+$$
and we note that $\psi_+$ is also chiral.
Since $\psi_+$ is chiral, we have used the fact 
${1\over 2}(1+\gamma_5)\psi_+=\psi_+$ in deriving the last equality above.

Now we use the rational approximation for $\epsilon(H_w)$ which we
write as
$$\epsilon(H_w) =c_oH_w+ \sum_{n=1}^N {c_n H_w\over H_w^2 + b_n }$$
Then
$$\eqalign{{\delta \epsilon(H_w)\over \delta U} & =
c_0 {\delta H_w\over \delta U} + \sum_{n=1}^N {c_n\over H_w^2+b_n} 
{\delta H_w\over \delta U} - \sum_{n=1}^N c_n 
{1\over H_w^2 + b_n} \bigl[ H_w {\delta H_w\over \delta U} +
{\delta H_w\over \delta U}H_w \bigr] {H_w\over H_w^2 + b_n} \cr
& = c_0 {\delta H_w\over \delta U} + \sum_{n=1}^N c_n {1\over H_w^2 + b_n}
\bigl[ b_n{\delta H_w\over \delta U} - H_w {\delta H_w\over \delta U} H_w 
\bigr] {1\over H_w^2 + b_n} \cr }$$

Using the expression for ${\delta \epsilon(H_w)\over \delta U}$, we write
the following expression for the derivative of the pseudo-fermion action.
$${\delta S_p\over \delta U} = -{1\over 2} (1-\mu^2)
\Biggl[ c_0 \psi^\dagger_+ {\delta H_w\over \delta U} \psi_+
+ \sum_{n=1}^N c_n {\psi_+^n}^\dagger \bigl[
b_n{\delta H_w\over \delta U} - H_w {\delta H_w\over \delta U} H_w \bigr]
\psi_+^n\Biggr]$$
where
$${1\over H_w^2 + b_n} \psi_+ = \psi^n_+$$

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